Problem: The line segment connecting $(-2,7)$ and $(3,11)$ can be parameterized by the equations
\begin{align*}
x &= at + b, \\
y &= ct + d,
\end{align*}where $0 \le t \le 1,$ and $t = 0$ corresponds to the point $(-2,7).$  Find $a^2 + b^2 + c^2 + d^2.$
Answer: Taking $t = 0,$ we get $(x,y) = (b,d) = (-2,7),$ so $b = -2$ and $d = 7.$

Taking $t = 1,$ we get $(x,y) = (a + b, c + d) = (3,11),$ so $a + b = 3$ and $c + d = 11.$  Hence, $a = 5$ and $c = 4.$

Then $a^2 + b^2 + c^2 + d^2 = 5^2 + (-2)^2 + 4^2 + 7^2 = \boxed{94}.$